# Definition of irreducible components in topological spaces

First of all the definition of irreducibility:

• A topological space $X$ is called reducible if $X$ can be written as a union of two non-empty and closed proper subsets of $X$. We call $X$ irreducible if it is not reducible.
• A subset $F$ of $X$ is called reducible resp. irreducible if it has this property in its subspace topology.

Now on Wikipedia I have found the following definition of irreducible component:

• An irreducible component of a topological space is a maximal irreducible subset.

My question: What do we mean with maximal here? I thought if a subset is irreducible it is automatically maximal because we can not split it up into two smaller closed sets. Can someone explain what is wrong with my thought? It would be great if someone could give me an specific example (e.g. from the Zariski topology).

I also have no idea why irreducible components are also closed but I hope that I will understand it if I could solve my problem above.

• If $A$ is irreducible then $\overline A$ is irreducible. For algebraic varieties probably irreducible component is the same as a closed irreducible subset. – user171326 Apr 23 '17 at 15:03

It's probably the case that $F \subset X$ is a maximal irreducible set if it's irreducible and the following implication holds: $$F \subset A \text{ and } A \text{ is irreducible} \implies F = A$$

Example: consider $X = \{0,1\}$ with the discrete topology. The set $\{1\}$ is an irreducible component.

Let $F$ be an irreducible component. Let's show that $\overline F$ is irreducible. Suppose not and write $\overline F = A \cup B$ where $A$ and $B$ are proper closed subsets of $\overline F$. We have $A = A' \cap \overline F$ and $B = B' \cap \overline F$ where $A'$ and $B'$ are closed in $X$. Let $A'' = A' \cap F$ and $B'' = B' \cap F$. We have that $A''$ and $B''$ are closed in $F$ and: $$A'' \cup B'' = (A' \cup B') \cap F = (A' \cup B') \cap \overline F \cap F = [(A' \cap \overline F) \cup (B' \cap \overline F)] \cap F = (A \cup B) \cap F = \overline F \cap F = F$$

Also note that if $A'' = F$, then $F \subset A'$, hence $\overline F \subset \overline A' = A'$, so $A = \overline F$, a contradiction. Similarly, we get that $A''$ and $B''$ are proper subsets of $F$. It's also evident that they're non-empty. Thus, $F$ is reducible, a contradiction. Therefore, $\overline F$ is irreducible and so, $F = \overline F$, so $F$ is closed.

Let $X$ be $\mathbb{N}$ in the cofinite topology, which has as only closed sets the finite sets and $X$ itself. This is hereditary: if $A \subseteq X$, its subspace topology is also cofinite.

Any infinite cofinite space $X$ is irreducible (we can only write it as $A \cup B$ both closed and not the whole space iff $A$ and $B$, and thus $X$ is finite.

So the even numbers in $X$ are irreducible but not maximally so (as $X$ itself is larger and also irreducible):

$A$ is an irreducible component of $X$ iff $A$ is irreducible and for every $A \subseteq B \subseteq X$: if $B$ is irreducible then $A = B$.

To check if a subset $F \subseteq X$ is irreducible, it is considered as its own topological space via the induced topology. So we isolate the set $F$, give it the induced topology (forgetting a lot of stuff about $X$!), and check if that new space is irreducible. But it might be the case that while $F$ considered as a space is irreducible, $F \cup G$ is also irreducible for some other subset $G \subseteq X$.

As an example, take the set $X = \{0, \ldots, n\}$ with the coarse topology where the only closed sets are $\emptyset$ and $X$. Here any subset $F \subseteq X$ will be coarse again, and hence irreducible, and so a maximal such subset is $X$. You can make a similar example using the Zariski topology: circle part of an irreducible curve on a page: that circled chunk should be irreducible as a topological space, but it is certainly not maximal.